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The margin of error is for the true average comprehensive strength :

\(E = {t_{\alpha /2}} \times \frac{s}{{\sqrt n }}\)

The margin of error is for the comprehensive strength of a single future specimen tested:

\(E = {t_{\alpha /2}} \times s\sqrt {1 + \frac{1}{n}} \)

Given that,

\(\begin{array}{l}n = 18\\\overline x = 64.41\\s = 10.32\\c = 98\% \end{array}\)

Determine the \(t\)-value by looking in the row starting with degrees of freedom \(\begin{array}{l}df = n - 1\\df = 18 - 1\\df = 7\end{array}\)

And in the column with \((1 - c)/2 = 0.01\) in the table of the student鈥檚 T distribution:

\({t_{\alpha /2}} = 2.567\)

The margin of error is then,

\(\begin{array}{l}E = {t_{\alpha /2}} \times \frac{s}{{\sqrt n }}\\ = 2.567 \times \frac{{10.32}}{{\sqrt {18} }}\\ \approx 6.2441\end{array}\)

The boundaries of the confidence interval, then become:

\(\begin{array}{l}\overline x - E = 64.41 - 6.2441\\\overline x - E = 58.1659\\\overline x + E = 64.41 + 6.2441\\\overline x + E = 70.6541\end{array}\)

Hence, the boundaries of the confidence interval is \((58.1659,70.6541)\).

Given that,

\({t_{0.02,17}} = 2.224\)

The critical value has been given as:

\({t_{\alpha /2}} = 2.224\)

The margin of error is then,

\(\begin{array}{l}E = {t_{\alpha /2}} \times s\sqrt {1 + \frac{1}{n}} \\ = 2.224 \times 10.32\sqrt {1 + \frac{1}{{18}}} \\ \approx 23.5806\end{array}\)

The lower bound of the prediction interval then become:

\(\begin{array}{l}\overline x - E = 64.41 - 23.5806\\\overline x - E = 40.8294\end{array}\)

Hence, The lower bound of the prediction interval is\(40.8294\).